Quiz 6 Flashcards

(36 cards)

1
Q

Comparison of oxygen solubility to other elements (3)

A

Oxygen is not very soluble in water (at max saturation, there is only ~14.6mg/L of oxygen in water) that’s 0.7g per kg

Some elements like CO2, NO, NH3, Cl2, and H2S are more soluble (NO, NH3, Cl2, and H2S are all very soluble and very toxic)
(NH3 is 900g per kg, and H2S is 7g per kg)

Other elements like CO, CH4, and N2 are less soluble (N2 is 0.3g per kg)

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2
Q

What is the equilibrium or saturation concentration (Cs) of a gas dissolved in a liquid a function of? (2)

A

The type of gas

The partial pressure of the gas in contact with the liquid

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3
Q

Henry’s Law for Dissolved Gases (4)

A

The solubility of a gas in a liquid depends on:

Temperature
The partial pressure of the gas at the surface of the liquid
The nature of the solvent
The nature of the gas

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4
Q

Partial pressure example (5)

A

When a can of Coke is bottled, it’s partial pressure is set at a high value

When the can is opened, the partial pressure of CO2 becomes much lower

The dissolved CO2 then bubbles out of the supersaturated liquid, escaping from the can

When the new low partial pressure equilibrium is established, the Coke will be flat

If the Coke is warm, the dissolved CO2 will be lost more quickly

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5
Q

What limits gas solubility?

A

Gas solubility is always limited by the equilibrium between the gas and saturated solution

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6
Q

Partial pressure and gas solubility (4)

A

When the temperature, nature of the solvent, and the nature of the gas remain constant, the concentration of dissolved gas depends on the partial pressure of the gas

This is because the partial pressure controls the number of gas molecule collisions with the surface of the solution

If the partial pressure is doubled, the number of collisions with the surface of the solution will double

This increase in number of collisions produces more dissolved gas

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7
Q

Two film theory of gas transfer equation (2)

A

Valid for quiescent and turbulent conditions

dc/dt = 1/V dm/dt = KLA/V(Cs-CL) = KLA(Cs-CL)

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8
Q

KL

A

Liquid film coefficient (m/hr)

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9
Q

A

A

Interfacial or absorbing

surface area of air (m2)

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10
Q

V

A

Volume of the liquid (m3)

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11
Q

CL

A

Average concentration of dissolved oxygen in bulk liquid (mg/L)

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12
Q

Cs

A

Saturation value of dissolved oxygen at the interface between liquid and air bubble (mg/L)

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13
Q

a

A

= A/V = the interfacial surface area of the air through with diffusion can occur generated by the particular aeration system per unit volume of water (m2/m3)

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14
Q

KLa (2)

A

Overall oxygen transfer coefficient

Takes into account velocity and area to volume

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15
Q

What are the 4 key factors that influence gas transfer in terms of the Two-Film Theory Equation?

A

KL
a
(Cs-CL)

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16
Q

What 7 principles do we need to consider when dissolving oxygen in water?

A

Salinity - solubility decreases with increasing salinity
Temperature - solubility decreases with increasing temperature
Pressure - solubility decreases with decreasing pressure
Turbulence - KL increases with increasing turbulence
Surface area - a increases with smaller bubbles as surface area increases
Concentration gradient - Cs increases with higher concentrations of oxygen
Average concentration - CL influenced by salinity, temperature and pressure

17
Q

Main list of factors influencing the transfer of dissolved oxygen into water (11)

A

Many more factors, but need to know for engineers, not us

Water temperature 
Depth of gas release
Contact time of gas bubble in water 
Size of gas bubble 
Rate of gas flow 
Type of diffuser 
Orifice diameter 
Oxygen concentration gradient 

Elevation - pressure
Turbulence
Salinity

18
Q

In lake and reservoir aeration we only need to know… (4)

A

How to manipulate the three core factors we can control

KL - the hydrodynamics of the system to influence KL (use a lot of turbulence)
a - the area of contact between the gas and liquid (use small bubbles)
(Cs-CL) - the concentration gradient between the gas and liquid phase (use pure oxygen if it makes economic and logistic sense)

19
Q

Measuring gas transfer (4)

A

There are 4 standard terms used to measure the transfer of oxygen into water:

KLa20 = overall oxygen transfer coefficient (per hour) 
SOTR = standard oxygen transfer rate (gO2/hr)
SAE = standard aeration efficiency (how much energy did you use?) (gO2 kW/hr)
SOTE = standard oxygen transfer efficiency (how much gas was transferred?) (%)
20
Q

Trade off with measuring gas transfer (2)

A

If one standard term to measure gas transfer is increased, it will likely decrease another term

Performance depends on what you are looking for - if increasing SAE is the goal, then the decreasing of SOTE won’t be a problem etc.

21
Q

Rule 1 for measuring gas transfer (2)

A

Rule 1: KLa and SOTR increase with flow rate and depth

SAE and SOTE increase with depth, decrease with flow rate

22
Q

Rule 2 for measuring gas transfer (2)

A

KLa, SOTR, SOTE increase with partial pressure

SAE decreases with increase in partial pressure

23
Q

Rule 3 for measuring gas transfer

A

KLa, SOTR, SAE, and SOTE increase with decrease in orifice size

24
Q

What are the 4 levers you have to influence oxygen transfer? (5)

A

Gas flow rate (which influences KL and a)

Depth of release (which influences Cs-CL)

Partial pressure of oxygen (which influences KL, a, and Cs-CL)

Orifice size (which influences KL and a)

All affect KLa(Cs-CL)

25
How do you resolve conflicting situation between achieving high KLa and STOR, and high SOTE and SAE? (3)
Use fine pore diffusers Use as many diffusers as possible Use high purity oxygen
26
Formula method for sizing destratification systems
Qw(x)=35.6C(x+0.8)sqrt(-Vo ln(1-(x/h+10.3))/ub) Where: ``` Qw(x)=water flow (m3/s) x=height above orifice (m) (depth) C=2Vo+0.05 Vo=air flow (m3/s at 1atm) h=depth of orifice (m) ub=25Vo+0.7 (m/s) ```
27
Field method for sizing destratification systems (3)
Use at least 1 SCFM air per 1,000,000 ft3 of lake volume Use at least 20 SCFM per 1,000,000 ft2 of lake surface area Doesn’t work on meromixes because of the shape of the lake basin, so just use the highest number or double the value to be safe
28
Morphogenic lake (2)
Predisposed to becoming meromictic because of their basin morphometry Ie. surface area small relative to depth like in Colomac Zone 2 Pit Lake, NWT
29
Colomac Mine story (3)
Royal Oak left without remediating mine site contaminated with tailings water full of cyanide, a chemical used to get gold out of ore Cleanup, maintenance etc. cost taxpayers millions of dollars What to do? Clean up areas and make companies pay for their own cleanup in the future
30
ENR Process (4)
The enhanced natural removal program involved the addition of mono-ammonium phosphate (MAP) to the Zone 2 Pit Mine to stimulate biological activity which removes cyanide and cyanide related compounds that are created from cyanide according to these reactions: 1. CN + H <=> HCN + metal cyanide complexes 2. CN + 1/2O2 + 3H2O ==> HCO3 + NH4 + OH 3. SCN+3H2O+2O2 =>SO4+NH4+HCO3+H
31
Zone 2 Pit Morphology (7)
High depth to length ratio Occurance of saline water Lack of natural outlet Protection from prevailing winds Rare mixing No oxygen in lower layers, high concentrations of NH3 and thiocyanate Under-ice circulation driven by salt freeze-out
32
ENR program and continued approaches to Colomac Zone 2 (4)
ENR program greatly reduced cyanide, but thiocyanate and ammonia remained in high concentrations in the monimolimnion Therefore, decided to remove rest of CN and thiocyanate, as well as ammonia released during these reactions, using O2 through artificial circulation using compressed air destratification system CN + 1/2O2 + 3H2O = HCO3 + NH4 + OH SCN + 3H2O + 2O2 = SO4 + NH4 + HCO3 + H
33
Where did all the oxygen go during Colomac Lake oxygenation? (3)
Was used up in reactions with cyanide and ammonia meaning the process was working After 30 days, thiocyanide was no longer detectable Ammonia was being replaced by ammonium so stayed a bit longer (H + NH3 <=> NH4) but continuing to add O2 to NH4 would eventually create nitrate (NO3) which is non-toxic
34
Bubbles in water equation (3)
Rise velocity of bubbles for 0.1cm to 2cm diameter explained by this equation: u = 1.02 sqrt(g re) Where: ``` u = terminal rise velocity (cm/sec) g = acceleration due to gravity (980cm/sec2) re = equivalent bubble radius (cm) ```
35
dc/dt and 1/V dm/dt
Partial change in concentration over time And partial change in mass per unit volume over time
36
What size of bubbles do you need to maximize KL? (2)
KL is at a maximum when bubbles are ~2-2.5mm (0.2-0.25cm) in diameter because this is the point of maximum rise velocity Therefore, overall rate of oxygen transfer (KLa) depends on size of bubbles, rise velocity, and area-volume ratio